How many cows are there? 
In a field, the grass increases in a constant rate. $17$ cows can eat all the grass in the field in $30$ days. $19$ cows can eat all the grass in $24$ days. Suppose,  a group of cows  started eating grass for $6$ days. Then $4$ cows are sold. The rest cows took $2$ more days to eat the remaining grass. What is the number of cows in the group?

My trying:
A cow eats a certain amount of grass in one day, call it $c$. 
The field grows by a certain amount each day, call it $g$. 
The field has some initial amount of grass: $i$ 
\begin{equation}
\begin{aligned}
i + 30g - 17\cdot30c &= 0\\
i+24g-19\cdot24c&=0
\end{aligned}
\end{equation}
Solving these two equations we get , $g = 9c$ . That means It takes 9 cows to eat one day's growth in one day.
 A: $$\begin{aligned}
V+30x&=30\times 17\times y\\
V+24x&=24\times 19\times y\\
V+8x&=6\times k\times y+2(k-4)\times y\\
V&=510y-30x\\
x&=9y\\
510y-22x&=(8k-8)y\\
510y-198y&=(8k-8)y\\
k&=40
\end{aligned}$$
where $V$ - value of grass on the field,
$x$ - speed of growth grass for a day,
$y$ - speed of 1 cow for a day,
$k$ - number of cows.
A: I'm going to set up some variables first.
Number of Cows: $C$
Amount of Grass on Day $n$: $G_n$
The rate of growth of grass: $x$ per day
The rate of consumption of grass: $y$ per day
So we have
$$G_{n+1}=G_n+x-Cy$$
which give
$$G_n=G_0+nx-nCy$$
The first two equations can be easily converted to
$$G_{30}=0=G_0+30x-30\cdot17\cdot y\tag{1}$$
$$G_{24}=0=G_0+24x-24\cdot19\cdot y\tag{2}$$
This gives
$$x=9y\tag{*}$$
For the last one, the recurrence relation no longer holds but the idea is the same, so we have
$$0=G_0+(6+2)x-6Cy-2(C-4)y$$
which simplifies to
$$0=G_0+8x-(8C-8)y\tag{3}$$
The rest is algebraic work. Using $(1)$ and $(*)$, you will get $C=40$.
A: Let
$i $ be the initial amount of grass on the field,
$a$ be the constant rate of growth of grass,
$b$ be the consumption rate of 1 cow for a day,
and $x $ be the number of cows in the situation asked
Given,
$$i+30a-17(30)b = 0 \tag1$$ 
$$i+24a-24(19)b = 0\tag2$$
Subtracting $(1)$ and $(2)$,
we get 
$$6a - 510b + 456b = 0$$
$$\Rightarrow 6a = 54b $$
or, $$a = 9b \tag3 $$ 
Putting this in $(1)$,
$$i+30a-17(30)b = 0 \tag1$$ 
$$ \Rightarrow i+30(9b)-17(30)b = 0$$ 
So,
$$i+270b-510b = 0$$ 
or
$$i = 240b \tag4$$
In the situation given,
$$i + 8a - 6xb = i_{2} \tag5$$
where $i_{2}$ is the amount of grass left before selling
Now we have $(x-4)$ cows left.
$$ i_{2} + 2a - 2(x-4)b = 0 \tag6$$
Combining $(5)$ and $(6)$ to eliminate $i_{2}$,
$$i + 8a - 6xb + 2a - 2(x-4)b = 0$$
Now using relations $(3)$ and $(4)$ we get,
$$(240b) + 8(9b) - 6xb + 2(9b) - 2(x-4)b = 0$$
Dividing both sides by $b$, we now get
$$240 + 54 - 6x + 18 - 2(x-4) = 0$$
$$\Rightarrow 312 - 6x  - 2x + 8 = 0$$
$$\Rightarrow 320 = 8x $$
so $x = 40$, or the initially there were 40 cows.
